Question

John is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than usual. When he weighs one of the sticks, he finds that it is 2.44 oz. The manufacturer's website states that the average weight of each stick is 2.00 oz with a standard deviation of 0.19 oz. Assume that the weight of the drumsticks is normally distributed. What is the probability of the stick's weight being 2.44 oz or greater? Give your answer as a percentage precise to at least two decimal places.

Answer 1

1.02% probability of the stick's weight being 2.44 oz or greater

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

`\mu = 2, \sigma = 0.19`

What is the probability of the stick's weight being 2.44 oz or greater?

As a decimal, this is 1 subtracted by the pvalue of Z when X = 2.44. So

`Z = (X - \mu)/(\sigma)`

`Z = (2.44 - 2)/(0.19)`

`Z = 2.32`

`Z = 2.32` has a pvalue of 0.9898

1 - 0.9898 = 0.0101

1.02% probability of the stick's weight being 2.44 oz or greater